1. Chapter 5 Class 10 Arithmetic Progressions
2. Serial order wise

Transcript

Ex 5.3 ,4 How many terms of the AP. 9, 17, 25 β¦ must be taken to give a sum of 636? AP is 9,17,25, β¦.. We know that Sum = π/2 (2a + (n β 1) d) Here, a = 9 d = 17 β 9 = 8 & Sum = Sn = 636 We need to find n Putting values in equation Sum = π/2 (2a + (n β 1) d) 636 = π/(2 ) (2 Γ 9 + (n β 1) Γ 8) 636 Γ 2 = n (18 + (n β 1) Γ 8) 1272 = n (18 + 8n β 8) 1272 = n (10 + 8n) 1272 = 10n + 8n2 0 = 10n + 8n2 β 1272 10n + 8n2 β 1252 = 0 2(5n + 4n2 β 636)= 0 5n + 4n2 β 636 = 0 4n2 + 5n β 636 = 0 We solve by quadratic formula Comparing equation with ax2 + bx + c = 0 Here, a = 4, b = 5, c = β 636 We know that, D = b2 β 4ac D = (5)2 β 4 Γ 4 Γ (β636) D = 25 + 10176 D = 10201 Hence, roots to equation are given by n = (β π Β± βπ·)/2π Putting values n = (β 5 Β± β10201)/(2 Γ 4) n = (β 5 Β± 101)/8 Solving So, n = 12 & n = (β53)/4 Since n cannot be negative, n = 12